Question

use synthetic division to find ((4x^{5}-4x^{4}-24x^{3}+6x^{2}-15x - 9)div(x - 3)). write your answer in the form (q(x)+\frac{r}{d(x)}), where (q(x)) is a polynomial, (r) is an integer, and (d(x)) is a linear polynomial. simplify any fractions.

Explanation:

Step1: Set up synthetic division

For dividing by \(x - 3\), we use \(3\) as the root. The coefficients of the dividend \(4x^{5}-4x^{4}-24x^{3}+6x^{2}-15x - 9\) are \(4, - 4, - 24, 6, - 15, - 9\).
Set up the synthetic division as:
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & & & & & \\ \hline & & & & & & \\ \end{array}$$

\]

Step2: Perform synthetic division

Bring down the first coefficient:
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & & & & & \\ \hline & 4 & & & & & \\ \end{array}$$

\]
Multiply \(4\) by \(3 = 12\), add to the next coefficient: \(-4+12 = 8\)
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & 12 & & & & \\ \hline & 4 & 8 & & & & \\ \end{array}$$

\]
Multiply \(8\) by \(3=24\), add to the next coefficient: \(-24 + 24=0\)
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & 12 & 24 & & & \\ \hline & 4 & 8 & 0 & & & \\ \end{array}$$

\]
Multiply \(0\) by \(3 = 0\), add to the next coefficient: \(6+0 = 6\)
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & 12 & 24 & 0 & & \\ \hline & 4 & 8 & 0 & 6 & & \\ \end{array}$$

\]
Multiply \(6\) by \(3=18\), add to the next coefficient: \(-15 + 18=3\)
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & 12 & 24 & 0 & 18 & \\ \hline & 4 & 8 & 0 & 6 & 3 & \\ \end{array}$$

\]
Multiply \(3\) by \(3 = 9\), add to the last coefficient: \(-9+9 = 0\)
\[

$$\begin{array}{r|rrrrrr} 3 & 4 & -4 & -24 & 6 & -15 & -9 \\ & & 12 & 24 & 0 & 18 & 9 \\ \hline & 4 & 8 & 0 & 6 & 3 & 0 \\ \end{array}$$

\]

The coefficients of the quotient polynomial \(q(x)\) are \(4,8,0,6,3\) and the remainder \(r = 0\). The degree of the quotient is one less than the dividend, so \(q(x)=4x^{4}+8x^{3}+0x^{2}+6x + 3=4x^{4}+8x^{3}+6x + 3\) and \(d(x)=x - 3\)

Answer:

\(4x^{4}+8x^{3}+6x + 3+\frac{0}{x - 3}\) or simply \(4x^{4}+8x^{3}+6x + 3\) (since the remainder is \(0\))